Question

For the hydrolysis of ATP at \(25^{\circ} \mathrm{C}(298 \mathrm{K})\) and \(\mathrm{pH} 7, \mathrm{ATP}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{ADP}+\mathrm{P}_{\mathrm{i}}+\mathrm{H}^{+},\) the standard free energy of hydrolysis \(\left(\Delta G^{\circ \prime}\right)\) is \(-30.5 \mathrm{kJ} \mathrm{mol}^{-1}\left(-7.3 \mathrm{kcal} \mathrm{mol}^{-1}\right),\) and the stan- dard enthalpy change \(\left(\Delta H^{\circ \prime}\right)\) is \(-20.1 \mathrm{kJ} \mathrm{mol}^{-1}\left(-4.8 \mathrm{kcal} \mathrm{mol}^{-1}\right)\) Calculate the standard entropy change \(\left(\Delta S^{\circ}\right)\) for the reaction, in both joules and calories. Why is the positive sign of the answer to be expected in view of the nature of the reaction? Hint: You may want to review some material from Chapter 1.

Short answer

The standard entropy change, \( \Delta S^{\circ} \, is 34.90 \text{J/(mol·K)} \) or 8.34 \text{cal/(mol·K)} \). The positive sign is expected due to increased disorder from ATP hydrolysis.

Step-by-step solution

Write Down Known Values

Note the given values of standard free energy change \(\Delta G^{\circ \prime} = -30.5 \text{kJ/mol} = -7.3 \text{kcal/mol}\) and standard enthalpy change \(\Delta H^{\circ \prime} = -20.1 \text{kJ/mol} = -4.8 \text{kcal/mol}\).

Recall the Relationship Between \( \Delta G \), \( \Delta H \), and \( \Delta S \)

The relationship between these thermodynamic quantities is given by the equation: \( \Delta G^{\circ \prime} = \Delta H^{\circ \prime} - T \Delta S^{\circ}\).

Solve for \( \Delta S^{\circ} \)

Rearrange the equation to solve for \( \Delta S^{\circ} \): \( \Delta S^{\circ} = \frac{\Delta H^{\circ \prime} - \Delta G^{\circ \prime}}{T}\).

Substitute the Known Values

Substitute \( \Delta H^{\circ \prime} \), \( \Delta G^{\circ \prime} \), and \( T = 298 \text{K} \) into the equation: \( \Delta S^{\circ} = \frac{-20.1 \text{kJ/mol} + 30.5 \text{kJ/mol}}{298 \text{K}}\).

Calculate \( \Delta S^{\circ} \) in Joules

Perform the calculation: \( \Delta S^{\circ} = \frac{10.4 \text{kJ/mol}}{298 \text{K}} = 0.03490 \text{kJ/(mol·K)} = 34.90 \text{J/(mol·K)}\).

Convert \( \Delta S^{\circ} \) to Calories

Since \( 1 \text{kJ} = 0.239006 \text{kcal} \), convert \( \Delta S^{\circ} \text{ in kJ/(mol·K)} \) to kcal/(mol·K): \( 34.90 \text{J/(mol·K)} \times 0.239006 \text{kcal/J} = 8.34 \text{cal/(mol·K)}\).

Explain the Positive Sign of \( \Delta S^{\circ} \)

A positive \( \Delta S^{\circ} \) is expected because the hydrolysis of ATP increases randomness: breaking one molecule of ATP into ADP and inorganic phosphate (Pi) results in more particles, increasing the disorder.

Key concepts

Standard Free Energy Change

The standard free energy change \( \Delta G^{\circ \prime} \) tells us how much energy is available to do work when a reaction occurs under standard conditions. For ATP hydrolysis, \( \Delta G^{\circ \prime} \) is \(-30.5 \text{kJ/mol} \.\) This negative value signifies that the reaction releases energy, making it exergonic. As ATP breaks down into ADP and Pi, the energy released can be used for various cellular processes.

The formula to relate free energy change, enthalpy change, and entropy change is:
\[ \Delta G^{\circ \prime} = \Delta H^{\circ \prime} - T \Delta S^{\circ} \]
This equation is central in thermodynamics as it helps predict whether a reaction will happen spontaneously. For ATP, the large negative \( \Delta G^{\circ \prime} \) explains why it's often called the

Standard Enthalpy Change

Standard enthalpy change \( \Delta H^{\circ \prime} \) measures the total heat exchange in a reaction under standard conditions. For ATP hydrolysis, \( \Delta H^{\circ \prime} \) is \( -20.1 \text{kJ/mol} \.\) This value tells us that the reaction releases heat, making it exothermic. When ATP is hydrolyzed, it not only releases free energy (\

Standard Entropy Change

Entropy change \( \Delta S^{\circ} \) reflects the change in disorder or randomness in a system. For ATP hydrolysis, we calculated \( \Delta S^{\circ} \) to be \( 34.90 \text{J/(mol·K)} \.\) This positive value indicates an increase in disorder as ATP breaks into ADP and Pi. In simpler terms, breaking one ATP molecule into more particles increases randomness, which is typical for breakdown reactions.

This aligns with the second law of thermodynamics, which states that systems naturally progress toward greater disorder. This positive \( \Delta S^{\circ} \) accompanies the energy release (

Thermodynamic Equations

Thermodynamic equations help us understand and predict chemical reactions. They relate key properties like enthalpy \( (\Delta H) \,\) entropy \( (\Delta S) \,\) and free energy \( (\Delta G) \). The primary equation we use in this context is:
\[ \Delta G = \Delta H - T \Delta S \]
In our exercise, by rearranging this equation, we calculated the entropy change:
\[ \Delta S^{\circ} = \frac{\Delta H^{\circ \prime} - \Delta G^{\circ \prime}}{T} \]
This formula helps find the balance between energy release (